Proving Abelian Property of (G,°) with f(a) = a^(-1)

[okeen]

Homework Statement

Let (G, °) be a group such that the mapping f from G into G defined by f(a) = a^(-1) is a homomorphism. Show that (G, °) is abelian.

The Attempt at a Solution

f(a) = a^(-1)
f(a^(-1)) = f(a)^(-1) = (a^-1)^-1 = a

in order for a group to be abelian it needs to meet the requirement a(i)*a(j) = a(j) * a(i)
° 1 a a^-1
1 1 a a^-1
a a a^2 1
a^-1 a^-1 1 a

since each side of the diagonal are the same then (G, °) is abelian.

 

[HallsofIvy]

You are assuming that G has only three members? Of course, every group containing 3 members is isomorphic to the rotatation group of a triangle which is abelian.

What if G contained six or more members?

 

[okeen]

I realize that my answer only takes into account for 3 members but I am having trouble coming up with a solution for all possible amounts of members

 

[HallsofIvy]

(ab)^-1= a^-1b^-1 if and only if a and b commute.

 

[rochfor1]

Interestingly enough, such a homomorphism must be an automorphism...not that I think it helps for this problem. HallsOfIvy is dead on with his hint.

 

[NoMoreExams]

(ab)^-1= a^-1b^-1 if and only if a and b commute.

So to follow up on that hint, see where f maps a, b and ab